Efield integral

this is the E field above a square loop with side=a at distance z on the z axis. by symmetry the Ex, Ey field cancel out. it really kind of bothers me that I can't see this substitution, including the sin(tan^-1(u)) portion. I guess it's an easy technique but I wonder how you get the substitution?

$$Ez=\frac{4\lambda z}{4\pi \varepsilon o}\int_{-a/2}^{a/2}\frac{ dx}{(z^2+x^2+a^2/4)^{3/2}}, x=\sqrt{a^2/4+z^2}tanu, dx=\sqrt{a^2/4+z^2}sec^{2}udu, I=\frac{1}{a^2/4+z^2}\int cosudu=\frac{1}{a^2/4+z^2}sinu\therefore Ez=\frac{8\lambda az}{4\pi\epsilon o \sqrt{2a^2+4z^2}z^2+a^2/4}$$

thanks!

arildno
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Thumb rule:
"To remove squares within square roots, use a suitable trig substitution. If stuff is more difficult than that, forget about it"

I try to do these by hand- I wonder if some people do all of these by hand?

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arildno
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I wonder if some people do all of these by hand?

Those who need to learn them, such as students. Those who are professionally competent don't, but at most, picks up a table of integrals. Only those are professionally competent who has learned them.

I'm not sure if maybe it's better to do a lot of problems out of mathematical and Schaum books. I had another career before this that was completely different than physics/mathematics
I've been looking at Apostol and Hubbard Calculus (introduces manifolds) and I'm not sure if I should study these since I've already done Stewart (problems are easy I know)

arildno
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I'm not sure what you're getting at.

$$sin(arctan(\frac{x}{(a^2/4+z^2)^{1/2}}))=\frac{x}{(a^2/4+x^2+z^2)^{1/2}}$$